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If $S\subseteq \kappa$ is stationary and $f(\alpha )\in A_{\alpha }$ for a kappa-filtration, then $f$ is constant on a stationary subset. Restrict to limit indices and use continuity to find a smaller stage containing each value. Fodor lemma fixes that stage on a stationary subset. Its size is less than $\kappa$, so club filter completeness makes one value fiber stationary.

The tree of restrictions ${\rho }_{\beta }\upharpoonright \alpha$, ordered by extension. For a club sequence on ${\omega }_2$ with club order types at most ${\omega }_1$, the Continuum hypothesis bounds each level by ${\aleph }_1$. Trace injectivity and Fodor lemma exclude a cofinal branch.

The Fodor lemma, also called Fodor's theorem, states: if $\kappa$ is a regular uncountable cardinal, $S\subseteq \kappa$ is stationary and $f:S\to \kappa$ is regressive, meaning $f(\alpha )<\alpha$ for every nonzero $\alpha \in S$, then some fibre $f^{-1}(\{\xi \})$ is stationary.

Delete zero from $S$, which preserves stationarity. Suppose no fibre is stationary. For each $\xi <\kappa$, choose a club set $C_{\xi }$ avoiding that fibre. The diagonal intersectionis a club. To prove unboundedness, start above an arbitrary bound and choose a strictly increasing sequence ${\alpha }_n$ with ${\alpha }_{n+1}$ in every $C_{\xi }$ for $\xi \le {\alpha }_n$. The intersection of these fewer-than-$\kappa$ clubs is club by club filter completeness. Put $\alpha ={\sup }_n{\alpha }_n<\kappa$. For any $\xi <\alpha$, every sufficiently late ${\alpha }_n$ lies in $C_{\xi }$, so its closure gives $\alpha \in C_{\xi }$. Thus $\alpha \in C$. For closure of $C$, let $\alpha <\kappa$ be a limit point of $C$; for each $\xi <\alpha$, the members of $C$ above $\xi$ witnessing that limit lie in $C_{\xi }$, and closure gives $\alpha \in C_{\xi }$.

Stationarity supplies $\alpha \in S\cap C$. Since $f(\alpha )<\alpha$, we have $\alpha \in C_{f(\alpha )}$, but that club was chosen to avoid the fibre containing $\alpha$. This contradiction proves the theorem. For completeness, club filter completeness follows by a similar diagonal sequence: cycle through fewer than $\kappa$ clubs over $\omega$ rounds, each time choosing a larger member of the next club. Regularity keeps the supremum below $\kappa$, and every club is visited cofinally often in the construction, so its closure contains the supremum. Their intersection is also closed.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal ultrafilter $U$ on $\kappa$ that is $\kappa$-complete: intersections of fewer than $\kappa$ members of $U$ remain in $U$. An elementary embedding $j:M\to N$ preserves truth of every first-order formula, with its parameters carried along by $j$.

The identity is always an elementary embedding of the universe into itself. The substantive question concerns a nonidentity embedding. In ZFC, the Kunen inconsistency theorem givesHere, as usual for class embeddings, restrictions to sets and the recursively defined critical sequence are available. We give the stationary-set proof, including its partition ingredient.

Suppose such a $j$ existed. It has a least moved ordinal $\kappa$, its critical point of an elementary embedding. To see why some ordinal must move, if all ordinals were fixed, induction on rank would fix every set: membership of $z$ in $j(x)$ is equivalent to membership of $z$ in $x$ once every lower-rank $z$ is fixed. A rank segment containing a moved set would then be fixed pointwise, a contradiction. The critical point is an uncountable cardinal. All finite ordinals and $\omega$ are definable and fixed. If $\kappa$ had cardinality $\mu <\kappa$, a bijection $f:\mu \to \kappa$ would be carried to a surjection $j(f):\mu \to j(\kappa )$ whose range still consists exactly of the fixed ordinals below $\kappa$, impossible since $j(\kappa )>\kappa$.

Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Applying $j$ to this Kunen critical sequence shifts it by one, so $j(\lambda )=\lambda$. Elementarity also gives $j({\lambda }^{+})={\lambda }^{+}$; write $\theta ={\lambda }^{+}$.

We need a stationary partition at a successor cardinal. If $S\subseteq {\lambda }^{+}$ is stationary, choose bijections $e_{\beta }:\lambda \to \beta$ for $\lambda \le \beta <{\lambda }^{+}$. For each $\xi <{\lambda }^{+}$, on the stationary tail $\{\beta \in S:\beta >\max (\lambda ,\xi )\}$ the function $\beta \mapsto e_{\beta }^{-1}(\xi )$ is regressive. The Fodor lemma supplies a stationary fiber $E_{\xi }$ with constant value $i_{\xi }<\lambda$. Some value $i$ occurs for ${\lambda }^{+}$ many $\xi$, since a union of $\lambda$ sets each of size at most $\lambda$ cannot have size ${\lambda }^{+}$. For distinct such $\xi$, their fibers are disjoint: $e_{\beta }(i)$ cannot equal two different ordinals. These fibers yield ${\lambda }^{+}$ disjoint stationary subsets; include any leftover points in one cell. Grouping cells also gives a partition into any nonzero number at most ${\lambda }^{+}$.

For completeness, the Fodor lemma used here follows from diagonal intersections of clubs. If a regressive function $f:S\to \theta$ had no stationary fiber, choose a club set $C_{\xi }$ disjoint from each fiber. The diagonal intersection $\{\beta <\theta :\forall \xi <\beta (\beta \in C_{\xi })\}$ is club by regularity and closure. A point $\beta$ in its intersection with $S$ would belong to $C_{f(\beta )}$, contradicting the choice of that club.

The set $S=\{\beta <\theta :\mathrm{cf}(\beta )=\omega \}$ is stationary: inside any club choose a strictly increasing countable sequence, whose supremum still lies below the regular uncountable $\theta$ and belongs to the club. Partition it as $⟨S_{\alpha }:\alpha <\kappa ⟩$. Its image $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ is another stationary partition of the same $S$.

The set $C=\{\beta <\theta :j‘‘\beta \subseteq \beta \}$ is club. Closure is immediate. To get a member above any given ordinal, repeatedly bound the images of all earlier ordinals and take a countable supremum; regularity of $\theta$ keeps each bound below $\theta$. Choose $\beta \in C\cap T_{\kappa }$. Since $\mathrm{cf}(\beta )=\omega$, a countable cofinal sequence in $\beta$ and $j(\omega )=\omega$ give $j(\beta )=\beta$: the image sequence remains below $\beta$ by closure under $j$, and $j(\beta )\ge \beta$.

But $\beta \in S_{\alpha }$ for some $\alpha <\kappa$. Applying $j$ and using $j(\alpha )=\alpha$ gives $\beta =j(\beta )\in T_{\alpha }$, contradicting its membership in the disjoint cell $T_{\kappa }$. This proves the assertion. The use of axiom of choice in the cardinal and stationary-partition arguments matters; the conclusion here is explicitly in ZFC.

For a nonzero limit ordinal $\delta$, a subset $C\subseteq \delta$ is a club set if it is unbounded in $\delta$ and contains every one of its limit points below $\delta$. A subset $S\subseteq \delta$ is stationary ifThis defines stationarity in a limit ordinal. The regular uncountable case is the usual stationary set setting; results such as Fodor lemma require that additional hypothesis, rather than an arbitrary limit ordinal.

There is a cofinal branch. We give a proof that does not require distinct limit-level nodes to have different predecessor chains.

The set $S=\{\alpha <\lambda :\mathrm{cf}(\alpha )=\kappa \}$ is stationary. Indeed a strictly increasing continuous $\kappa$-sequence in any club set has supremum in that club set, below $\lambda$, of cofinality $\kappa$. At each $\alpha \in S$, there are fewer than $\kappa$ pairs of nodes on $T_{\alpha }$. For every pair whose predecessor chains below $\alpha$ differ, choose a height where they differ. Since $\mathrm{cf}(\alpha )=\kappa$, all these heights are bounded by some ${\beta }_{\alpha }<\alpha$. Consequently nodes in $T_{\alpha }$ having the same predecessor at ${\beta }_{\alpha }$ have identical predecessor chains below $\alpha$.

The Fodor lemma states that a regressive function on a stationary subset of a regular uncountable cardinal is constant on a stationary subset. Applying it here, ${\beta }_{\alpha }$ has a constant value $\beta$ on a stationary subset $S^{\prime }$. Choose $t_{\alpha }\in T_{\alpha }$ for each $\alpha \in S^{\prime }$. The level $T_{\beta }$ has fewer than $\kappa <\lambda$ nodes. Partitioning $S^{\prime }$ according to the predecessor of $t_{\alpha }$ at height $\beta$, one fiber $S^{\prime \prime }$ is stationary, since the union of fewer than $\lambda$ nonstationary sets is nonstationary. Let its common predecessor be $t$.

For $\alpha <{\alpha }^{\prime }$ in $S^{\prime \prime }$, the predecessor of $t_{{\alpha }^{\prime }}$ at level $\alpha$ and $t_{\alpha }$ have the same predecessor $t$ at level $\beta$. Their chains below $\alpha$ therefore agree. For each $\nu <\lambda$, choose $\alpha \in S^{\prime \prime }$ above $\nu$ and let $b_{\nu }$ be the predecessor of $t_{\alpha }$ at level $\nu$. The preceding comparison makes $b_{\nu }$ independent of that choice. The nodes $b_{\nu }$ form a chain through every level:This proves the uniformly narrow regular-height tree branch theorem. The uniform bound below the smaller regular $\kappa$ is stronger than merely bounding each level below $\lambda$.

Let $E=\{\alpha <\kappa :\mathrm{cf}(\alpha )=\omega \}$. This set is stationary: in any club set, choose a strictly increasing countable sequence and take its supremum, which lies in the club set and has cofinality $\omega$. For each $\alpha \in E$ choose an increasing cofinal sequence $c_{\alpha }:\omega \to \alpha$.

Fix $\beta <\kappa$. For every $\alpha \in E$ above $\beta$, some $c_{\alpha }(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary sets is nonstationary, because fewer than $\kappa$ club sets have club filter completeness, so some cell is stationary. On it the regressive function $\alpha \mapsto c_{\alpha }(n)$ has, by Fodor lemma, a stationary fiber at a value $\gamma >\beta$.

Let $I_n=\{\gamma <\kappa :\{\alpha \in E:c_{\alpha }(n)=\gamma \}\text{ is stationary}\}$. The preceding argument says ${\bigcup }_n{I}_n$ is unbounded in $\kappa$. Since $\mathrm{cf}(\kappa )>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint stationary sets. Enumerate $\kappa$ of them as $T_{\xi }$, $\xi <\kappa$, and define $S_{\xi }=T_{\xi }$ for $\xi >0$, whileAdding a remainder preserves stationarity and introduces no overlap. ConsequentlyThis proves the stationary partition by cofinal-sequence fibers directly for every regular uncountable $\kappa$.

A kappa-filtration is an increasing continuous sequence with union $A$ and $|A_{\alpha }|<\kappa$ at every stage. Intersect $S$ with the club set of nonzero limit ordinals. For each remaining $\alpha$, continuity gives $A_{\alpha }={\bigcup }_{\beta <\alpha }{A}_{\beta }$, so choose $r(\alpha )<\alpha$ with $f(\alpha )\in A_{r(\alpha )}$. This is a regressive function. Fodor lemma gives a stationary subset $S_1$ and a fixed $\beta$ such that all these values lie in $A_{\beta }$.

Since $|A_{\beta }|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a club set avoiding each one. Their intersection is club set by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, andThis proves the filtration form of Fodor lemma; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.

Take $\kappa ={\omega }_1$ and choose a club sequence on ${\omega }_2$, each $C_{\delta }$ of order type at most ${\omega }_1$. For a successor $\delta$ use its predecessor as a singleton; at a limit use a cofinal sequence of minimal length. Form the minimal-walk treeordered by proper extension. Its height is ${\omega }_2$.

For $\xi <\delta$, the initial segment $C_{\delta }\cap \xi$ has order type strictly below ${\omega }_1$, and is countable. The strict inequality follows because a point of $C_{\delta }$ at or above $\xi$ occurs later in its enumeration. Hence every entry of every trace is countable. Under the Continuum hypothesis, for $|\alpha |\le {\aleph }_1$,There are at most ${\aleph }_1$ finite sequences of such sets. The trace coherence lemma for minimal walks says that, for $\beta >\alpha$, the value ${\rho }_{\beta }(\alpha )$ determines ${\rho }_{\beta }\upharpoonright \alpha$. The case $\beta =\alpha$ adds at most one node. Thus $|T_{\alpha }|\le {\aleph }_1$ for every level.

Suppose that $T$ had a cofinal branch, and take the union of its functions, $f$, with domain ${\omega }_2$. Every ${\rho }_{\beta }$ is injective by the proper-initial-segment argument, so $f$ is injective too. On the stationary setthis set is stationary because the supremum of a strictly increasing ${\omega }_1$-sequence from any club set has cofinality ${\omega }_1$ and lies in that club. The union of the finitely many countable entries of $f(\alpha )$ is bounded below $\alpha$. Assign a strict upper bound below $\alpha$ to obtain a regressive function. By Fodor lemma, there is a stationary $S^{\prime }\subseteq S$ and a single $\eta <{\omega }_2$ such that every entry of $f(\alpha )$ lies inside $\eta$ for $\alpha \in S^{\prime }$. There are at most ${\aleph }_1$ such finite sequences by the same cardinal arithmetic, but $|S^{\prime }|={\aleph }_2$, contradicting injectivity.

Therefore $T$ is an aleph-two Aronszajn tree:

Fodor lemma says that if $\lambda$ is a regular uncountable cardinal, $S\subseteq \lambda$ is stationary, and $f:S\to \lambda$ is regressive, thenHere a regressive function satisfies $f(\alpha )<\alpha$ at every nonzero $\alpha \in S$; removing $0$ has no effect on stationarity. Both regularity and stationarity are part of the hypotheses.

A function on a set of ordinals with $f(\alpha )<\alpha$ at every nonzero point of its domain. On a stationary subset of a regular uncountable cardinal, Fodor lemma makes it constant on a stationary subset.

For any nonzero limit ordinal $\delta$, a club set is an unbounded subset closed under limit points below $\delta$. A subset is stationary if it meets every such club set. Statements concerning the completeness of the club filter or Fodor lemma require regular uncountable height; they should not be transferred to arbitrary singular or countable limit ordinals without checking their hypotheses.

Every stationary subset of a successor cardinal ${\lambda }^{+}$ can be partitioned into ${\lambda }^{+}$ disjoint stationary subsets. For each $\beta \in [\lambda ,{\lambda }^{+})$ choose a bijection $e_{\beta }:\lambda \to \beta$. For each $\xi <{\lambda }^{+}$, the function $\beta \mapsto e_{\beta }^{-1}(\xi )$ is regressive on the stationary tail with $\beta >\xi$. The Fodor lemma gives a stationary fiber with one value $i_{\xi }<\lambda$. Some value occurs for ${\lambda }^{+}$ many indices, whose fibers are pairwise disjoint. Assign any leftover part of $S$ to one fiber.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.

For infinite regular cardinals $\kappa <\lambda$, a $\lambda$-tree whose levels all have size less than $\kappa$ has a cofinal branch. At levels of cofinality $\kappa$, bound all heights distinguishing distinct predecessor chains. The Fodor lemma makes this bound constant on a stationary set; one predecessor at that height occurs stationarily often. The resulting predecessor chains agree below their common heights and unite into a branch through every level. Distinct limit-level nodes with identical predecessor chains do not affect this argument.